1. Chapter 9 Stoichiometry
Lots of math !! 

2. Unit Essential Question:
What numerical information can we find from a balanced chemical reaction?

3. Lesson Essential Question:
How can we tell the amounts of substances involved in a reaction?

4. Section 1: Calculating Quantities in Reactions
Coefficients of balanced equations show proportions (ratios) of substances in reactions.
Similar to amounts in a recipe.
Use this information to adjust amount of product(s) made or reactants needed.
Relative amounts (coefficients) in equation = moles.
Example: 2 C8H O2  16 CO H2O
2 moles of C8H18 react with 25 moles of O2 to form 16 moles of CO2 and 18 moles of H2O
Like we discussed in Chapter 8, an equation can tells us amounts. These amounts can be utilized to find the amount produced in grams or other units. We can even work backwards and find out how we need to start with to make a certain amount of product.

5. Stoichiometry
Deals with quantities of substances in chemical reactions.
Compare amounts (moles) and also masses.
Limiting reactants and percent yield
can be calculated.
Can be useful for everyday activities such as baking.
We will learn 3 basic types of problems. We will look at some real-world examples as well. One good one, not mentioned by your book, is the meals used by the military when they do not have access to a kitchen.

6. Mole Ratio Needed in order to perform calculations!
Allows you to convert between substances.
Set up using fractions and conversion factors.
Use dimensional analysis.
Get ratios from balanced equation.
Ex: 2H2 + O2  2H2O
Mole ratio examples: 2mol H2/1mol O2,
1mol O2/2mol H2O, or 2molH2/2molH2O
A very important concept of this chapter is the mole ratio. Every single problem will have a mole ratio in it. You need the balanced equation to find the mole ratio.

7. Steps to Converting Between Amounts of Substances
Step 1: Balance the chemical equation.
Step 2: Convert given amount to moles.
Remember- always start with your given!
Step 3: Use a mole ratio from the balanced equation to convert from the amount of one substance to the amount of another substance.
Use dimensional analysis for all conversions!
Step 4: Convert out of moles if asked for grams or the number of particles.

8. Sample Problem #1
Consider the reaction for the commercial preparation of ammonia:
N2 + 3 H2  2 NH3
How many moles of hydrogen are needed to prepare 312 moles of ammonia?
Notice that mole ratios convert between substances and units stay the same (all moles)!
Moles of nitrogen needed?

9. Sample Problem #2
What mass of NH3 can be made from 1,221 g of H2 and excess N2?
N2 + 3 H2  2 NH3

10. Sample Problem #3
How many grams of C5H8 form from 1.89 x molecules C5H12?
C5H12 (l)  C5H8 (l) + 2 H2 (g)

11. Lesson Essential Question:
How can we calculate the amount of products made in a reaction?

12. Section 2: Limiting Reactants and Percent Yield
Limiting reactant – substances that limit the amount of product that can be formed.
Limiting reactants run out first!
Excess reactant – substances that are not used up in the reaction. There are some left over.
Do not limit the amount of product that can be formed.
Think back to the sandwich activity- what were...
LR [limiting reagent(s)]?
ER [excess reagent(s)]?
I like to use the abbreviations LR and ER for limiting reactant and excess reactant. Feel free to use them on your homework, quizzes, labs, tests, etc.

13. Theoretical & Actual Yield
This is what we have been calculating in practice problems so far!
Theoretical & Actual Yield
Theoretical: the amount that should be able to be produced based on the limiting reactant.
If everything in the reaction went according to plan, and all of the reactant(s) reacted, this is how much product should be made.
This is NOT the same as the actual yield- amount that is produced based on an experiment
Error occurs, so actual yield is less than the theoretical yield.
Can never be greater than the theoretical yield.
The amount you calculate is the theoretical in this case and the actual is what you get when you do an experiment (lab).

14. Determining the Limiting Reactant
You will be given amounts of two or more reactants.
Convert all amounts to moles.
Use mole ratios in the balanced chemical equation to determine the number of moles of a product that will be produced.
Convert all reactant amounts into the same product!
Convert the moles of product to grams of product.
Do this for each reactant amount given and see how much product each produces.
Whichever produces LESS product is your limiting reactant- it will be completely consumed first.

15. Sample Problem #5
Identify the limiting reactant and the theoretical yield of phosphorus acid, H3PO3, if 225 g of PCl3 is mixed with 123 g of H2O
PCl3 + 3 H2O  H3PO3 + 3 HCl

16. Sample Problem #6
Determine the limiting reactant if 14.0g N2 are mixed with 9.0g H2. What is the theoretical yield in grams of the product?
N2 + 3H2  2NH3

17. Application Cheapest ingredient = excess reactant.
Most expensive = limiting reactant.
Cost effective!
Examples:
Cider vinegar
ER = oxygen; LR = ethanol
Banana Flavoring
ER = acetic acid; LR = isopentyl alcohol
In industry, executives would try to keep the most expensive ingredient as the limiting reactant. This is ideal so none is left over and wasted.

18. (actual yield/theoretical yield) x 100 = % yield
Percent Yield
Actual yield is always less than the theoretical yield.
LR is not always 100% used up.
Can be due to human errors, not reacting to completion, reverse reactions and/or unwanted side reactions occurring, and loss during purification.
Efficiency is another name for percent yield.
Percent Yield:
(actual yield/theoretical yield) x 100 = % yield
If your percent yield is greater than 100%, there could be unreacted or unwanted materials in your product.
It is not always human error that causes the percent yield to be less than 100%; the side reactions will cause less product to form than theoretically possible. You should not get more than 100% - if you do, then there were side reactions or unreacted materials that are mixed with the product.

19. Sample Problem #7
Using your answers from Sample Problem #6, calculate the percent yield if 16.1 g NH3 are formed in an experiment.
N2 + 3 H2  2 NH3

20. Sample Problem #8
How many grams of CH3COOC5H11 should form if 4808 g are theoretically possible and the percent yield for the reaction is 80.5%?